Find a basis for the kernel of the given Linear transformation. Let $n$ be a positive integer and let $V$ be the space of real polynomials whose degree does not exceed $n$. Consider the map
$f: V\mapsto V$, $f(P(X))= P(X)+ (1-X)P'(X)$ where $P'(X)$ is the derivative of $P$.
Give a basis for the kernel of $f$.
[I got confused with all the terms involved in the solution, is my claim correct?]
Claim :$Ker(f) = 0$
Let $P(X)= a_nx^n + a_{n-1}x^{n-1} +...+a_2x^2 + a_1x+a_0$.
$P'(X) = na_nx^{n-1}+ (n-1)a_{n-1}x^{n-2} +....a_1$
For $f(P(X))=0(X)$, where $0(X)$ is a zero polynomial. which further implies $P(X) +(1-X)P'(X) =0$ i.e $P(X)=(X-1)P'(X)\implies a_nx^n + a_{n-1}x^{n-1} +...+a_2x^2 + a_1x+a_0 =na_nx^{n}+ [(n-1)a_{n-1} -na_n]x^{n-1} + [(n-2)a_{n-2} -(n-1)a_{n-1}]x^{n-2}....[a_1-2a_2]x - a_1$
Comparing coefficients we have $a_n = na_n,a_{n-1}=(n-1)a_{n-1} -na_n,...,a_1= a_1-2a_2$ and $a_1 = a_0$
if $n\neq 1$ then all the coefficients equals $0$ otherwise $P(X) = a(X-1)$ for some $a\in \mathbb {R}$ therefore $span(X-1)= Ker(f)$
 A: Observe that by the comment by Bungo, $\;Q(x)\in\ker f\implies Q(x)=ax+b\,,\,\,a,b\;$ scalars, and then if we assume that $\;Q(x)=ax+b\in\ker f\;$ , we get
$$0=f(ax+b)=(ax+b)+(1-x)a=a+b\implies b=-a\implies$$
$$Q(x)=ax-a=a(x-1)\in\text{Span}\,\{1-x\}:=\left\{\;p(x)\in V\;|\;p(x)=k(1-x)\;,\;\;k\in\Bbb R\;\right\}$$
Thus, $\;\ker f=\text{Span}\,\{1-x\}\;$ and we obtain  $\;\;\dim\ker f=1\;$ ...
Another way: Suppose we have
$$p(x)\in\ker f\iff f(x)=-(1-x)p'(x)$$
the last equality is a polynomial one and it thus remains true under any substitution of $\;x\;$ . We thus substitute
$$x=1\implies p(1)=-(1-1)p'(1)=0\implies p(1)=0\stackrel{\text{Residue theorem for pols.}}\iff $$
$$\iff(1-x)\mid p(x)\iff p(x)=(1-x)q(x)\,,\,\,\text{for some}\;\;q(x)\in V$$
and we thus get that $\;p(x)\;$ is a polynomial multiple of $\;1-x\;$. We now apply, once again, the observation made by Bungo in the comments and get that since $\;\deg p(x)\le 1\;$ then it must be $\;\deg q=0\iff q(x)=a\in\Bbb R\;$ , is a scalar...and we can finish as above.
A: The solution to $$\frac{p'(x)}{p(x)}=\frac{1}{x-1}$$
is easily computed: $c(x-1)$.
